Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 0-32198-238-X
ISBN 13: 978-0-32198-238-4

Chapter 1 - Linear Equations in Linear Algebra - 1.8 Exercises - Page 69: 5

Answer

$\begin{bmatrix}3-3k\\1-2k\\k\end{bmatrix}$ for any real number $k$ (not unique)

Work Step by Step

Finding the vector with a given image under some matrix transformation is equivalent to finding the solution set of a linear system. Hence, we begin with the augmented matrix: $\begin{bmatrix}1&-5&-7&-2\\-3&7&5&-2\end{bmatrix}$ Add $3$ times row (1) to row (2): $\begin{bmatrix}1&-5&-7&-2\\0&-8&-16&-8\end{bmatrix}$ Multiply row (2) by $-\frac{1}{8}$: $\begin{bmatrix}1&-5&-7&-2\\0&1&2&1\end{bmatrix}$ Add $5$ times row (2) to row (1): $\begin{bmatrix}1&0&3&3\\0&1&2&1\end{bmatrix}$ Since the third column contains no pivot, the solution is not unique, and we have $\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}=\begin{bmatrix}3-3k\\1-2k\\k\end{bmatrix}$.
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